Quantum Computation (Lecture QC-3: Quantum Algorithms) Lu s Soares - - PowerPoint PPT Presentation

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Quantum Computation (Lecture QC-3: Quantum Algorithms) Lu s Soares - - PowerPoint PPT Presentation

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Quantum Computation (Lecture QC-3: Quantum Algorithms) Lu s Soares Barbosa MFES - Arquitectura e C alculo May 2019 Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover A quantum machine Structure of a quantum

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Quantum Computation

(Lecture QC-3: Quantum Algorithms)

Lu´ ıs Soares Barbosa MFES - Arquitectura e C´ alculo

May 2019

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

A quantum machine

Structure of a quantum algorithm

  • 1. State preparation (fix initial setting): typically the qubits in the

initial classical state are put into a superposition of many states;

  • 2. Transform, through unitary operators applied to the superposed

state;

  • 3. Measure, i.e. projection onto a basis vector associated with a

measurement tool.

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

My first quantum program

Is f : 2 − → 2 constant, with a unique evaluation?

Oracle

|x⟩ Uf |x⟩ |y⟩ |y ⊕ f (x)⟩

where ⊕ stands for exclusive disjunction.

  • The oracle takes input |x, y to |x, y ⊕ f (x)
  • for y = 0 the output is |x, f (x)
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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

My first quantum program

Is f : 2 − → 2 constant, with a unique evaluation?

Oracle

  • The oracle is a unitary, i.e. reversible gate

|x⟩ Uf |x⟩ Uf |x⟩ |y⟩ |y ⊕ f (x)⟩ |y⟩

|x, (y ⊕ f (x)) ⊕ f (x) = |x, y ⊕ (f (x) ⊕ f (x)) = |x, y ⊕ 0 = |x, y

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

My first quantum program

Idea: Avoid double evaluation by superposition

|x⟩ Uf |1⟩ H

  • The circuit computes:
  • utput = |x|0 ⊕ f (x) − |1 ⊕ f (x)

√ 2 =

  • |x |0−|1

√ 2

⇐ f (x) = 0 |x |1−|2

√ 2

⇐ f (x) = 1 = (−1)f (x) |x|0 − |1 √ 2

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

My first quantum program

Idea: Avoid double evaluation by superposition

|0⟩ H Uf H

  • |1⟩

H

(H ⊗ I) Uf (H ⊗ H)(|01) Input in superposition |σ1 = |0 + |1 √ 2 |0 − |1 √ 2 = |00 − |01 + |10 − |11 2

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

My first quantum program

|σ2 = (−1)f (0)|0 + (−1)f (1)|1 √ 2 |0 + |1 √ 2

  • =

   (+1)

  • |0+|1

√ 2 |0−|1 √ 2

  • ⇐ f constant

(+1)

  • |0−|1

√ 2 |0−|1 √ 2

  • ⇐ f not constant

|σ3 = H|σ2 =    (+1) |0

  • |0−|1

√ 2

  • ⇐ f constant

(+1) |1

  • |0−|1

√ 2

  • ⇐ f not constant

To answer the original problem is now enough to measure the first qubit: if it is in state |0, then f is constant.

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

The Deutsch-Jozsa Algorithm

Generalizing Deutsch’s algorithm to functions whose domain is an initial segment n of N, encoded into a binary string (i.e. the set of natural numbers from 0 to 2n − 1. Assuming f : 2n − → 2 is either balanced or constant, determine which is the case with a unique evaluation

Oracle

|x⟩ /n Uf /n |x⟩ |y⟩ | f (x) ⊕ y⟩

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Using H⊗n to put n qubits superposed

Computing H⊗n

H = 1 √ 2 (−1)0∧0 (−1)0∧1 (−1)1∧0 (−1)1∧1

  • H⊗2 =

1 √ 2 (−1)0∧0 (−1)0∧1 (−1)1∧0 (−1)1∧1

  • ⊗ 1

√ 2 (−1)0∧0 (−1)0∧1 (−1)1∧0 (−1)1∧1

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Using H⊗n to put n qubits superposed

Computing H⊗n

H⊗2 = 1 √ 2

  • (−1)0∧0

(−1)0∧1 (−1)1∧0 (−1)1∧1

  • ⊗ 1

√ 2

  • (−1)0∧0

(−1)0∧1 (−1)1∧0 (−1)1∧1

  • =

1 2     (−1)00,00 (−1)00,01 (−1)01,00 (−1)01,01 (−1)00,10 (−1)00,11 (−1)01,10 (−1)01,11 (−1)10,00 (−1)10,01 (−1)11,00 (−1)11,01 (−1)10,10 (−1)10,11 (−1)11,10 (−1)11,11     where x, y = (x0 ∧ y0) ⊕ (x1 ∧ y1) ⊕ · · · ⊕ (xn ∧ yn) Note that (−1)a∧b ⊗ (−1)a ′∧b ′ = (−1)a∧a ′⊕b∧b ′ = (−1)aa ′,bb ′

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Using H⊗n to put n qubits superposed

Computing H⊗n

In general, the value of H⊗n at coordinates i, j (row and column numbers as binary strings) is given by H⊗n

i,j

= 1 √ 2n (−1)i,j Applying H⊗n to an arbitrary basic state |i (which is a column vector with 1 in line i and 0 everywhere else), extracts the i-column of H⊗n: H⊗n|i = 1 √ 2n

  • x∈{0,1}n

(−1)x,i|x e.g. H⊗2|0 = 1 2     1 1 1 1 1 −1 1 −1 1 1 −1 −1 1 −1 −1 1         1     = 1 2     1 1 1 1     = 1 2

  • x∈{0,1}n

|x

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

First move: Uf (I ⊗ H)|x, 1

|x⟩ /n Uf /n

  • |1⟩

H ⇑ |ϕ0⟩ ⇑ |ϕ1⟩ ⇑ |ϕ2⟩ (6.65)

|ϕ1 = |x|0 − |1 √ 2 = |x, 0 − |x1 √ 2 |ϕ2 = |x|f (x) ⊕ 0 − |f (x) ⊕ 1 √ 2 = (−1)f (x)|x|0 − |1 √ 2

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Second move: (H⊗n ⊗ I)Uf (H⊗n ⊗ H)|0, 1

Put input |x into a superposition in which all 2n possible strings have equal probability: H⊗n|0.

|0⟩ /n H⊗n /n Uf /n H⊗n /n |1⟩ H ⇑ |ϕ0⟩ ⇑ |ϕ1⟩ ⇑ |ϕ2⟩ ⇑ |ϕ3⟩

|ϕ1 =

  • x∈{0,1}n |x

√ 2n |0 − |1 √ 2 |ϕ2 =

  • x∈{0,1}n (−1)f (x)|x

√ 2n |0 − |1 √ 2

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Second move: (H⊗n ⊗ I)Uf (H⊗n ⊗ H)|0, 1

|0⟩ /n H⊗n /n Uf /n H⊗n /n |1⟩ H ⇑ |ϕ0⟩ ⇑ |ϕ1⟩ ⇑ |ϕ2⟩ ⇑ |ϕ3⟩

|ϕ3 =

  • x∈{0,1}n (−1)f (x)

z∈{0,1}n (−1)z,x|z

√ 2n |0 − |1 √ 2 =

  • x,z∈{0,1}n (−1)f (x)(−1)z,x|z

√ 2n |0 − |1 √ 2 =

  • x,z∈{0,1}n (−1)f (x)⊕z,x|z

√ 2n |0 − |1 √ 2

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Finally: observe!

When do the top qubits of |ϕ3 collapse to |0?

Making |z = |0 (and thus z, x = 0 for all x) leads to |ϕ3 =

  • x∈{0,1}n (−1)f (x)|0

√ 2n |0 − |1 √ 2 i.e. the probability of collapsing to |0 depends only on f (x)

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Finally: observe!

Analyse the top qubits

f is constant at 1

  • x∈{0,1}n (−1)|0

√ 2n

=

−(2n)|0 2n

= −|0 f is constant at 0

  • x∈{0,1}n 1|0

√ 2n

=

(2n)|0 2n

= |0 f is balanced

  • x∈{0,1}n (−1)f (x)|0

√ 2n

=

0|0 2n

= 0|0 because half of the x will cancel the other half The top qubits collapse to |0 only if f is constant Exponential speed up: f was evaluated once rather than 2n − 1 times

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Search problems

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Search problems

A more precise formulation

Given a function f : 2n − → 2 such that there exsits a unique binary string x∗ st f (x) =

  • 1

⇐ x = x∗ ⇐ x = x∗ determine e.

A quadratic speed up

  • Worst case for a classic algorithm: 2n evaluations of f
  • Worst case for Grover’s algorithm:

√ 2n evaluations of f

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Grover’s algorithm

Oracle Uf inverts the phase at |x∗

Recall from Deutsch-Josza:

|x⟩ /n Uf /n

  • |1⟩

H ⇑ |ϕ0⟩ ⇑ |ϕ1⟩ ⇑ |ϕ2⟩ (6.65)

|ϕ2 = (−1)f (x)|x|0 − |1 √ 2 =

  • −|x |0−|1

√ 2

⇐ x = x∗ +|x |0−|1

√ 2

⇐ x = x∗

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Grover’s algorithm

Oracle Uf inverts the phase at |x∗

Thus, providing as input a balanced superposition of all possible states, via H⊗n|0, the oracle is able to detect the solution and shift its phase:

  • However, the probability of collapsing to |x∗ is equal to the one of

collapsing to any other basic state becase | − 1 √ 2n |2 = | − 1 √ 2n |2

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Boosting the phase separation

The trick: Inversion around the mean

  • e ′ = mean + (mean − e) ⇔ e ′ = −e + 2mean

Computing the mean (example)

      

1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5 1 5

             53 38 17 23 79       =       42 42 42 42 42      

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Boosting the phase separation

The trick: Inversion around the mean

For A the grid matrix, V ′ = −V + 2AV = (−I + 2A)V multiplying any state by (−I + 2A) inverts amplitudes around the mean.

Healthiness test

Operator (−I + 2A) is unitary, because

  • (−I + 2A)† = (−I + 2A)
  • (−I + 2A) (−I + 2A) = I − 2A − 2A + 4A2 = I − 4A + 4A = I
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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Combining effects over time to amplify the right phase

Example

  • Start with [10, 10, 10, 10, 10]T
  • Invert the fourth entry: [10, 10, 10, −10, 10]T
  • Invert around mean (6): [2, 2, 2, 22, 2]T

Note 22 − 2 = 20

  • Invert the fourth entry again: [2, 2, 2, −22, 2]T
  • Invert around mean (−2.8): [−7.6, −7.6, −7.6, 16.4, −7.6]T

Note 16.4 + 7.6 = 24.

  • ...

The right phase is amplified in successive iterations

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Combining effects to amplify the right phase

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Combining effects to amplify the right phase

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Grover’s algorithm

Repeat √ 2n times Phase inversion Inversion about mean |0⟩ /n H⊗n /n Uf /n −I + 2A /n

  • |1⟩

H ⇑ ⇑ ⇑ ⇑

  • Questions
  • Why

√ 2n iterations?

  • How to implement the oracle?
  • Generalizations?

e.g. multiple search requires

  • 2n

t iterations for t the multiplicity

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Grover’s algorithm is everywhere

SAT (= Boolean satisfiability) problems

Determining values for Boolean variables so that a given Boolean expression evaluates to true

  • NP-complete
  • Many problems, like scheduling, can be converted into a SAT
  • Can be seen as a search problem whose goal is to find a precise

combination of Boolean values that yields true

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Second thoughts

Creating a uniform superposition of all basis states does not allow to satisfactorily solve NP-complete problems Let Uf encode a SAT formula on n Boolean variables: Uf (|i ⊗ |0) = |i ⊗ |f (i) Applying Uf to a superposition obtained via H⊗n|0, which evaluates the truth assignment of all possible binary strings, will return a binary string that satisfies the formula iff the last qubit has value 1 after the measurement, and this happens with a probability that depends on the number of binary assignments that satisfy the formula (e.g.

τ 2n , for τ such assignments).

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Quantum Algorithms The Deutsch-Jozsa Algorithm Quantum Search: Grover

Second thoughts

Although, in general, solving NP-hard problems in polynomial time with quantum computers is probably not possible (cf P = NP?), there is a recipe to produce faster equivalent quantum algorithms:

  • Create a uniform superposition of basis states
  • Make the basis states interact with each other so that the modulus
  • f the coefficients for some (desirable) basis states increase, which

implies that the other coefficients decrease.

  • How to do it ... depends on the problem